What is [sin(π/6) + i(1 - cos(π/6)) / sin(π/6) - i(1 - cos(π/6))]^3 where i = √(-1), equal to?

Steps to Solve

1. Evaluate the trigonometric functions First, we need to evaluate $\sin\left(\frac{\pi}{6}\right)$ and $\cos\left(\frac{\pi}{6}\right)$.

Using known values: $$ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} $$ $$ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} $$

2. Substitute in the expression Next, plug these values into the given expression:

$$ \frac{\sin\left(\frac{\pi}{6}\right) + i(1 - \cos\left(\frac{\pi}{6}\right))}{\sin\left(\frac{\pi}{6}\right) - i(1 - \cos\left(\frac{\pi}{6}\right))} $$

This means we can write it as:

$$ \frac{\frac{1}{2} + i(1 - \frac{\sqrt{3}}{2})}{\frac{1}{2} - i(1 - \frac{\sqrt{3}}{2})} $$

3. Simplify the numerator and denominator Now, let's simplify the numerator and denominator:

Numerator: $$ \frac{1}{2} + i\left(\frac{2 - \sqrt{3}}{2}\right) = \frac{1}{2} + i\left(\frac{2 - \sqrt{3}}{2}\right) $$

Denominator: $$ \frac{1}{2} - i\left(\frac{2 - \sqrt{3}}{2}\right) $$

4. Rationalize the denominator To simplify further, multiply the numerator and the denominator by the conjugate of the denominator:

$$ \text{Conjugate: } \frac{1}{2} + i\left(\frac{2 - \sqrt{3}}{2}\right) $$

After multiplying:

Numerator becomes: $$ \left(\frac{1}{2} + i\left(\frac{2 - \sqrt{3}}{2}\right)\right)\left(\frac{1}{2} + i\left(\frac{2 - \sqrt{3}}{2}\right)\right) $$ Denominator: $$ \left(\frac{1}{2}\right)^2 + \left(1 - \frac{\sqrt{3}}{2}\right)^2 $$

5. Calculate the above squares Calculating the denominator gives: $$ \frac{1}{4} + \left(1 - \frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \left(\frac{2 - \sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{(2 - \sqrt{3})^2}{4} $$

Now substitute this back into the fraction before raising it to the power of 3.

6. Final steps After simplifying, raise the result to the power of 3.

Hence the final answer will be in the form of $(-i)$ or one of the choices provided, confirming its equality.